The set of open segments joining an open set with a point is open

Question

Let $U\subseteq\mathbb{R}^n$ be an open set and $x\in\mathbb{R}^n\setminus U$.

For every $u\in U$ let $L_u$ be the line segment joining $u$ and $x$, without the points $u$ and $x$.

Let $V=\cup\{L_u:u\in U\}$.

Is the set $V$ open?

My idea is that every point $v\in V$ should have a corresponding point $u_v\in U$ such that $u_v$ "projects" a "shrunk" neighborhood of $v$ in $V$. I can't formalize this idea yet.

Any hints?

Answer 1

Take $v \in V$. It exists $u \in U$ such that $v \in L_u$. Which means that it exists $\lambda \in (0,1)$ such that $v = (1-\lambda) x +\lambda u$.

Now as $U$ is supposed to be opened, an open ball $B(u, r)$ of radius $r>0$ and centered on $u$ is included in $U$.

Denote by $f_\lambda(y)=x+\lambda (y-x)$ the homothetic transformation centered on $x$ with ratio $\lambda$. The image of $B(u,r)$ under $f_\lambda$ is $B(v,\lambda r)$ which implies $B(v,\lambda r) \subseteq V$, proving that $V$ is open.